What is the volume of the solid produced by revolving #f(x)=1/x-1/x^2, x in [2,6] #around the x-axis?

1 Answer
bp
Dec 18, 2016

#49/324#

Explanation:

The portion of the graph of f(x) in the interval [2,6] looks like in the figure below:
enter image source here
If this portion is revolved around x-axis, the volume of the solid so generated can be worked out as follows:

Consider an element of the area enclosed by the x-axis and the curve, at a distance x from the origin and length y= f(x). If this is rotated about x-axis a circular disc of area #pi y^2# If width of this elementary disc is dx, its volume would be #pi y^2 dx#. The volume of the entire solid would be

#int_2^6 pi y^2 dx= int_2^6 pi (1/x- 1/x^2)^2 dx#

=#pi int_2^6 (1/x^2 -2/x^3 +1/x^4)dx#

=#pi[-1/x +1/x^2 -1/(3x^3)]_2^6#

=#pi[-1/6 +1/36 -1/648 +1/2 -1/4 +1/24]=49/324#

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